Unit 3 · Topic 3.6

3.6 Sinusoidal Function Transformations

In f(θ) = a sin(b(θ + c)) + d, each constant does one job: a sets the amplitude, b sets the period, c shifts the graph left or right, and d moves the midline. The same rules work for cosine. You'll read these off equations and graphs and build equations from features.

Key terms

phase shift
vertical shift
horizontal dilation
period = 2π/|b|
amplitude = |a|

The general form

Sinusoidal functions can be written as f(θ) = a sin(b(θ + c)) + d or g(θ) = a cos(b(θ + c)) + d, where a ≠ 0 and b ≠ 0. These are the transformations from topic 1.12 applied to sine and cosine.

Constant	Effect	Result
a	Vertical dilation by a factor equal to the absolute value of a; reflection over the midline if a < 0	Amplitude = absolute value of a
b	Horizontal dilation by a factor of 1 over the absolute value of b	Period = 2π divided by the absolute value of b
c	Horizontal shift (phase shift) by −c	Left c units if c > 0, right if c < 0
d	Vertical shift by d	Midline y = d

Reading each piece

Amplitude: |a|. The maximum is d + |a| and the minimum is d − |a|.

Period: 2π/|b|. A larger |b| squeezes more cycles into the same space, so the period gets shorter. For sin(3θ), the period is 2π/3.

Phase shift: −c. In sin(θ − π/4), c = −π/4, so the graph moves π/4 to the right.

Midline: y = d.

To move a single point: a point (θ, y) on y = sin θ moves to (θ/b − c, a · y + d) on y = a sin(b(θ + c)) + d. Divide the input by b and then subtract c; multiply the output by a and then add d.

For example, the peak (π/2, 1) of sin θ moves to (π/4 − π/4, 3 · 1 + 2) = (0, 5) on y = 3 sin(2(θ + π/4)) + 2.

Factor before you read the shift

If the inside isn't factored, like sin(2θ + π), factor out b first: sin(2(θ + π/2)). Now c = π/2, so the shift is π/2 to the left, not π.

Another reliable way: a sine cycle starts where its inside equals 0. Solve 2θ + π = 0 to get θ = −π/2, the same answer.

Building an equation from a graph

Find the max and min to get d = (max + min)/2 and |a| = (max − min)/2.

Find the period (max to max) and solve period = 2π/b for b.

Pick a convenient feature for the shift. With cosine, use a maximum: a cosine graph with a > 0 starts at a peak. With sine, use a midline crossing where the graph is rising.

Many different equations describe the same graph. For example, cos θ = sin(θ + π/2) = −cos(θ + π). Any correct one earns credit.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Reading the features
For f(θ) = −3 cos(2(θ − π/4)) + 1, find the amplitude, period, phase shift, midline, maximum and minimum, and describe f at θ = π/4.
Show the solutionHide the solution
1. Step 1: a = −3: amplitude 3, with a reflection over the midline.
2. Step 2: b = 2: period 2π/2 = π.
3. Step 3: c = −π/4: phase shift π/4 to the right.
4. Step 4: d = 1: midline y = 1, so the maximum is 1 + 3 = 4 and the minimum is 1 − 3 = −2.
5. Step 5: At θ = π/4 the inside is 0, and cos 0 = 1, so f(π/4) = −3(1) + 1 = −2. Because of the reflection, the shifted cycle starts at a minimum instead of a maximum.
Answer: Amplitude 3, period π, shift π/4 right, midline y = 1, max 4, min −2; f has a minimum of −2 at θ = π/4.
Example 2
Equation from a graph
A sinusoidal graph has a maximum at (1, 8), and the next minimum is at (4, 2). Write an equation for it.
Show the solutionHide the solution
1. Step 1: Midline: d = (8 + 2)/2 = 5. Amplitude: (8 − 2)/2 = 3.
2. Step 2: From a max to the next min is half a period, so the period is 2(4 − 1) = 6. Then 2π/b = 6 gives b = π/3.
3. Step 3: Use cosine, which starts at a maximum, shifted so the maximum is at x = 1: f(x) = 3 cos((π/3)(x − 1)) + 5.
4. Step 4: Check the minimum: f(4) = 3 cos(π) + 5 = −3 + 5 = 2.
Answer: f(x) = 3 cos((π/3)(x − 1)) + 5 (other equivalent forms are fine).
Example 3
Trap: shifting before factoring
What is the phase shift of g(θ) = sin(2θ + π/3)?
Show the solutionHide the solution
1. Step 1: The trap answer is “π/3 to the left,” read straight from the parentheses.
2. Step 2: Factor out 2: g(θ) = sin(2(θ + π/6)). The shift is π/6 to the left.
3. Step 3: Check: the cycle starts where 2θ + π/3 = 0, at θ = −π/6.
Answer: π/6 to the left.

Common mistakes

Using b as the period. The period is 2π/|b|, so sin(4θ) has period π/2, not 4.
Reading the shift from an unfactored inside like sin(3θ − π). Factor: 3(θ − π/3), a shift right π/3.
Shifting the wrong way: sin(θ + c) moves the graph left when c > 0.
Giving a negative amplitude. The amplitude is |a|; the sign of a only tells you about a reflection.

On the exam

Expect to match an equation to a graph or to write an equation from a graph or table. The periodic-modeling free-response question often asks for the values of a, b, c and d.
On no-calculator questions, keep b exact, such as π/6, rather than a decimal.

Connected topics

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